Recent content by honestrosewater

None?

The zero function z sends all points to the same point, say, r? So composition is only commutative if every other function also sends r to r? I feel totally lost. But then z(a) = 1/2 and f(a) = x/2 would work as examples for your A?

How can you compose two functions from the same hom-set? Am I missing something that requires the domain and codomain to be equal?

Homework Statement Produce an example in the category in which objects are open sets in R2 and morphisms are continuous maps, to illustrate that Hom sets need not be abelian groups in this category. The Attempt at a Solution I'm missing something stupid here. I'd think that I was...

If x is an arbitrary real number, is x necessarily 3? No. Is x possibly 3? Yes. This is not a difficult question if you think about it. Perhaps you want to hear that the truth value is variable. It depends on the value assigned to x. The truth value of a formula containing variables is a...

Did you try starting with your conclusion and working backwards? What rules allow you to infer a disjunction? Or put it into another form and see what your next-to-last step would need to be.

Yes, correct, this is what you need to prove. Why must every model that satisfies S also satisfy α? You need a proof. You can give a syntactic argument that S syntactically implies α. You can also give a semantic argument that no counterexample can exist. The syntactic argument seems easier.

Producing one model doesn't prove it for all models. There are structures that satisfy both the original S and α. The nonnegative integers are such a structure. But this doesn't mean that all structures satisfying one will satisfy the other, as you proved with your counterexample. You need...

So are you more interested in classifying the signal (as one of your four words) or in processing the signal into something useful to a classifier? If you want to classify it, I think you'll have the most luck doing it statistically with an HMM. This has been the predominant approach for long...

Sure, you can do it many ways. If you don't have a formal system, you can think about it this way. 0) ((A v B) & (~A v C)) -> (B v C) (0) is an implication formula (i.e., its topmost operator is implication). An implication is only false when the antecedent is true and the consequent is...

Your new S needs to imply ∀x(odd(x) ⊃ even(s(x))). You need to prove what you said, that any interpretation that satisfies S also satisfies a. You can do this by deriving ∀x(odd(x) ⊃ even(s(x))) from S. This would be a simple matter if you also had ∀x[~odd(x) <-> even(x)] and ∀x[~even(x) <->...

Okay, what part do you understand? Do you know any other formal proof systems? Can you show by any means that this formula is always true? ((A v B) & (~A v C)) -> (B v C)

Just match up the complements. You have several. I suppose it makes sense for a function to become a variable or a constant in unification, and if unifying two functions, the higher arity one becomes the lower arity (which is a generalization of the first rule). 1) ~E(x) v V(x) v C(f(x)) 2)...

So you want the standard interpretation here, with "0" as 0 and "s" as successor? It doesn't hurt to say so. Can you derive ∀x(odd(x) ⊃ even(s(x))) from your new set? Just state the proof. You only need to know two things. The easiest way to make a theorem derivable from a set of axioms is...

The axioms for S generate an infinite alternating sequence of odd and even objects that starts with 0. So every even object is followed by an odd object. To not model a, you need to have an odd object that is not followed by an even one. Since the sequence following 0 alternates infinitely, you...